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 Post subject: On Riemann SurfacesPosted: Fri Nov 13, 2015 12:59 am
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Joined: Tue Nov 10, 2015 8:25 pm
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• Suppose that $X$ is a connected and compact Riemann surface and let $\displaystyle f : X \longrightarrow \mathbb{C}$ be a holomorphic function. Show that $\displaystyle f$ is constant.
• Let $\displaystyle f : \mathbb{C} \longrightarrow \mathbb{C}$ be a holomorphic bounded function. Show that there is a unique (holomorphic) extension $\displaystyle \hat{f} : \hat{ \mathbb{C} } \longrightarrow \hat{ \mathbb{C} }$ and then conclude Liouville's theorem (regarding $\mathbb{C}$).

Note that $\displaystyle \hat{ \mathbb{C} } = \mathbb{C} \cup \{ \infty \}$.

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 Post subject: Re: On Riemann SurfacesPosted: Sun Mar 13, 2016 2:51 pm
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Joined: Tue Nov 10, 2015 8:25 pm
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(1) Suppose that $f$ is not constant. As a non-constant holomorphic map between Riemann Surfaces, $f$ is open, so $f(X)$ is an open subset of $\mathbb{C}$. Since $X$ is compact and $f$ is continuous, $f(X)$ is also compact and thus a closed subset of $\mathbb{C}$. But $\mathbb{C}$ is connected, so $f(X) = \mathbb{C}$ as $f$ is not constant. This means that $\mathbb{C}$ is compact! Contradiction.

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 Post subject: Re: On Riemann SurfacesPosted: Sun Mar 13, 2016 3:01 pm
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Joined: Tue Nov 10, 2015 8:25 pm
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(2) Consider the function $g$ given by $g(z) = f(\frac{1}{z})$ on $\mathbb{C} \smallsetminus \{0\}$. Clearly, $g$ is holomorphic on $\mathbb{C} \smallsetminus \{0\}$ and bounded on a neighborhood of $0$, due to the boundedness assumption on $f$. By Riemann's Removable Singularities Theorem, $0$ is a removable singularity of $g$, which means that $g$ can be defined at $0$ in such a way that it becomes holomorphic on all of $\mathbb{C}$. It follows that $\hat{f} \ \colon \hat{\mathbb{C}} \longrightarrow \mathbb{C} \ , \ \hat{f}(z) = \begin{cases} f(z) \ , \ z \in \mathbb{C} \\ g(0) \ , \ z = \infty \end{cases}$is a holomorphic extension of $f$ to $\hat{\mathbb{C}}$. By (1), $\hat{f}$ is constant and consequently $f$ itself is constant.

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